Answer:
The percentage of the weight supported by the front wheel is A= 19.82 %
Explanation:
From the question we are told that
The center of gravity of the plane to its nose is [tex]z = 2.58 m[/tex]
The distance of the front wheel of the plane to its nose is [tex]l = 0.800\ m[/tex]
The distance of the main wheel of the plane to its nose is [tex]e = 3.02 \ m[/tex]
At equilibrium the Torque about the nose of the airplane is mathematically represented as
[tex]mg (z- l) - G_B *(e - l) = 0[/tex]
Where m is the mass of the airplane
[tex]G_B[/tex] is the weight of the airplane supported by the main wheel
So
[tex]G_B =\frac{mg (z-l)}{(e - l)}[/tex]
Substituting values
[tex]G_B =\frac{mg (2.58 -0.8 )}{(3.02 - 0.80)}[/tex]
[tex]G_B = 0.8018 mg[/tex]
Now the weight supported at the frontal wheel is mathematically evaluated as
[tex]G_F = mg - G_B[/tex]
Substituting values
[tex]G_F = mg - 0.8018mg[/tex]
[tex]G_F = (1 - 0.8018) mg[/tex]
[tex]G_F = 0.1982 mg[/tex]
Now the weight of the airplane is = mg
Thus percentage of this weight supported by the front wheel is [tex]A = 0. 1982 *100 =[/tex] 19.82 %
